Circum radius of a `DeltaABC` is 3 units, let O be the circum and H be the orthocentre then the value of `(1)/(64)(AH^(2)+BC^(2))(BH^(2)+AC^(2))(CH^(2)+AB^(2))` equals :

Options

A. `3^(4)`
B. `9^(3)`
C. `27^(6)`
D. `81^(4)`

Answer

To solve the problem, we need to evaluate the expression:

\[ \frac{1}{64} (AH^2 + BC^2)(BH^2 + AC^2)(CH^2 + AB^2) \]

Given that the circumradius \( R \) of triangle \( ABC \) is 3 units, we can start by using the relationships between the sides, angles, and the circumradius.

### Step 1: Define the sides and angles
Let:
- \( BC = a \)
- \( AC = b \)
- \( AB = c \)

### Step 2: Express \( AH^2, BH^2, CH^2 \) in terms of \( R \)
Using the known relationships in triangle geometry, we have:
- \( AH = 2R \cos A \)
- \( BH = 2R \cos B \)
- \( CH = 2R \cos C \)

Thus:
\[ AH^2 = (2R \cos A)^2 = 4R^2 \cos^2 A \]
\[ BH^2 = (2R \cos B)^2 = 4R^2 \cos^2 B \]
\[ CH^2 = (2R \cos C)^2 = 4R^2 \cos^2 C \]

### Step 3: Express \( BC^2, AC^2, AB^2 \) in terms of \( R \)
Using the sine rule, we can express the sides in terms of the circumradius:
- \( BC = a = 2R \sin A \)
- \( AC = b = 2R \sin B \)
- \( AB = c = 2R \sin C \)

Thus:
\[ BC^2 = (2R \sin A)^2 = 4R^2 \sin^2 A \]
\[ AC^2 = (2R \sin B)^2 = 4R^2 \sin^2 B \]
\[ AB^2 = (2R \sin C)^2 = 4R^2 \sin^2 C \]

### Step 4: Substitute into the expression
Now substituting these values into the original expression:

\[ AH^2 + BC^2 = 4R^2 \cos^2 A + 4R^2 \sin^2 A = 4R^2 (\cos^2 A + \sin^2 A) = 4R^2 \]

Similarly:
\[ BH^2 + AC^2 = 4R^2 (\cos^2 B + \sin^2 B) = 4R^2 \]
\[ CH^2 + AB^2 = 4R^2 (\cos^2 C + \sin^2 C) = 4R^2 \]

### Step 5: Combine the results
Now substituting back into the expression:

\[ (AH^2 + BC^2)(BH^2 + AC^2)(CH^2 + AB^2) = (4R^2)(4R^2)(4R^2) = 64R^6 \]

### Step 6: Final expression
Now substituting this into our original expression:

\[ \frac{1}{64} (64R^6) = R^6 \]

### Step 7: Substitute \( R = 3 \)
Since \( R = 3 \), we have:

\[ R^6 = 3^6 = 729 \]

### Conclusion
Thus, the final value of the expression is:

\[ 729 \]

### Options
The options provided were:
- A) \( 3^4 \)
- B) \( 9^3 \)
- C) \( 27^6 \)
- D) \( 81^4 \)

Since \( 729 = 9^3 \), the correct answer is:

**Option B: \( 9^3 \)**